(a) The table shows some values of \(y = \frac{6}{x}.\)
| x | -5 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 5 |
| y | -1.2 | -1.5 | 6 | 2 | 1.5 | 1.2 |
(i) Complete the table.
(ii) On the grid, draw the graph of \(y = \frac{6}{x}\) for -5 ≤ x ≤ -1 and 1 ≤ x ≤ 5.
(iii) On the same grid, draw the line y = 4.
(iv) Find the co-ordinates of the point where the line y = 4 crosses the graph of \(y = \frac{6}{x}\).
(b)
(i) On this grid, plot the point A (–1, –3).
(ii) Draw a line with gradient 2 through point A.
(iii) Write down the equation of your line in the form y = mx + c.
▶️ Answer/Explanation
(a)(i) Ans:
| x | -5 | -4 | -3 | -2 | -1 | 1 | 2 | 3 | 4 | 5 |
| y | -1.2 | -1.5 | -2 | -3 | -6 | 6 | 3 | 2 | 1.5 | 1.2 |
Compute \(y = \frac{6}{x}\) for missing \(x\) values. For example, when \(x = -3\), \(y = \frac{6}{-3} = -2\). Similarly, fill all gaps.
(ii) Plot the points and draw two smooth hyperbola curves (one in the negative \(x\) region and one in the positive \(x\) region).
(iii) Draw a horizontal line at \(y = 4\).
(iv) Ans: (1.4 to 1.6, 4)
Solve \(\frac{6}{x} = 4\) to get \(x = 1.5\). Thus, the intersection point is \((1.5, 4)\).
(b)(i) Plot the point A at \((-1, -3)\).
(ii) Draw a line through A with slope 2 (i.e., for every 1 unit right, go 2 units up).
(iii) Ans: \(y = 2x – 1\)
Using point-slope form: \(y + 3 = 2(x + 1)\) simplifies to \(y = 2x – 1\).
(a)
(i) Complete the table of values for \( y = x^2 + x \).
(ii) On the grid, draw the graph of \( y = x^2 + x \) for \(-3 \leq x \leq 3\).
(iii) On the grid, draw the line \( y = 10 \).
(iv) Use both your graphs to solve \( x^2 + x = 10 \) for \(-3 \leq x \leq 3\).
(b) Another line, \( L \), has the equation \( y = \frac{2}{3}x – 5 \).
(i) Write down the gradient of \( L \).
(ii) Write down the equation of a straight line that is parallel to \( L \).
(c)
Write the equation of the line, \( K \), in the form \( y = mx + c \).
▶️ Answer/Explanation
(a)
(i) Ans: For \( x = 1 \), \( y = 2 \); for \( x = 2 \), \( y = 6 \); for \( x = 3 \), \( y = 12 \).
Substitute \( x \) into \( y = x^2 + x \):
\( x = 1 \): \( 1^2 + 1 = 2 \),
\( x = 2 \): \( 4 + 2 = 6 \),
\( x = 3 \): \( 9 + 3 = 12 \).
(ii) Plot the points \((-3,6), (-2,2), (-1,0), (0,0), (1,2), (2,6), (3,12)\) and draw a smooth curve.
(iii) Draw a horizontal line at \( y = 10 \).
(iv) Ans: \( x \approx 2.7 \) (from intersection of \( y = x^2 + x \) and \( y = 10 \)).
(b)
(i) Ans: Gradient \( = \frac{2}{3} \).
(ii) Ans: Any line of the form \( y = \frac{2}{3}x + c \), where \( c \neq -5 \).
(c)
Ans: \( y = 2x – 3 \).
From the graph, the gradient \( m = \frac{\text{rise}}{\text{run}} = 2 \), and the y-intercept \( c = -3 \).